too, but this is not true in general.

Let us introduce the following zero-order di¬erential operator

P ± = (j ’ n)±, ± ∈ Aj (M ).

¨

3.6. HODGE THEORY ON KAHLER MANIFOLDS 119

Apparently [P, L] = 2L because L increases the degree of a di¬erential

form by 2; similarly [P, Λ] = ’2Λ. Also we have

LEMMA 3.6.11

[L, Λ] = P.

Proof. Since it is a purely punctual statement, we can again use the

same arguments as in Theorem 1.14 to reduce the problem to the case

M = Cn with the standard K¨hler form. Then if we view A— (Cn ) as

a

ˆ ˆ

the completed tensor product A— (C)— · · · —A— (C), we can write

L = L1 —Id— · · · —Id + Id—L2 —Id · · · + · · ·

where Lj is the operator L on the j-th copy of C. By adjunction

we have similar decompositions for Λ and for P . Therefore we only

have to prove the statement in the case of C with the K¨hler form

a

√

ω = ’ ’1dz§d¯/2. Here we may check √ statement directly. On

z the

√

0-forms we have ΛL(1) = Λ(ω) = 2 ’1(’ ’1/2) = 1 and P (1) = ’1

as desired. On 1-forms both √ vanish identically. Finally, on 2-forms

sides

we have LΛ(dz§d¯) = L(2 ’1) = dz§d¯ and P (dz§d¯) = dz§d¯,

z z z z

thus we are done.

Now if one looks closely on the relations [L, Λ] = P , [P, L] = 2L,

and [P, Λ] = ’2Λ then one recognizes the relationship between the

operators Λ, L, and P and the algebra sl(2, R). The latter is the algebra

of traceless real 2 — 2 matrices with basis

10 01 00

H= ,X = , and Y = ,

0 ’1 00 10

which have exactly the same commutation relations: [H, X] = 2X,

[H, Y ] = ’2Y , and [X, Y ] = H. Thus the 3 dimensional Lie subalgebra

Lef spanned by Λ, L and P of the algebra Op(A— (M )) of all operators

is isomorphic to sl(2, R) via the correspondence H ’ P , X ’ L, and

Y ’ Λ. In this way the algebra A— (M ) of real-valued di¬erential forms

on M becomes a sl(2, R)-module. Because of the fact that the Laplacian

∆ commutes with L, Λ, and P one sees that the space of harmonic forms

H— (M ) ‚ A— (M ) is an sl(2, R)-submodule. Note that a vector space V

is an sl(2, R)-module if we have three operators H, X, Y on V satisfying

120 CHAPTER 3. HODGE THEORY

the above bracket relations. It follows that it is useful for us to study

¬nite-dimensional modules over sl(2, C) (also called representations of

sl(2, C)). Let V be a ¬nite-dimensional vector space acted upon by

operators X, Y and H as above. We will prove

THEOREM 3.6.12 For any integer m ≥ 0, there is exactly one (up

to isomorphism) irreducible representation Vm of sl(2, C) of dimension

m + 1.

We can describe Vm explicitly. For V0 we take just a trivial 1-dimensional

sl(2, C)-module. Let V1 = C2 with the standard action of sl(2, C), and

let Vm = S m C2 be the m-th symmetric power of V1 . The action of

sl(2, R) is extended to an action on the symmetric powers by the usual

rules for the action of a derivation on a product.

To get a di¬erent and in a way more conceptual model, let Pm

be the space of homogeneous polynomials in two variables x and y of

‚ ‚ ‚ ‚

degree m. Also, let H = x ‚x ’ y ‚y , X = x ‚y , and Y = y ‚x . It is

an easy exercise to check that we get the right commutation relations.

It is possible to ¬nd a basis in Pm consisting of eigenvectors of H,

namely, if vi = xi y m’i , then Hvi = (2i ’ m)vi for 0 ¤ i ¤ m. We have

Xvi = (m ’ i)xi+1 y m’i’1 = (m ’ i)vi+1 , and similarly Y vi = ivi’1 .

The next important result is the so-called complete reducibility of

¬nite-dimensional representations of sl(2, C).

THEOREM 3.6.13 Any ¬nite-dimensional representation V of sl(2, C)

decomposes into a direct sum

V Vm1 • · · · • Vmk

of the irreducible representations described above.

We start proving these results with a simple lemma

LEMMA 3.6.14 Let V be a ¬nite-dimensional sl(2, C)-module. If u ∈

V is such that Xu = 0, and H m u = 0 for some m ≥ 1, then Hu =

Y u = 0.

¨

3.6. HODGE THEORY ON KAHLER MANIFOLDS 121

Proof. We start with the formula

[X, Y k ] = k(H + k ’ 1)Y k’1 ,

which is easy to deduce by induction. Indeed, for k = 1 it is clear, and

if it is true for k = k0 then we have

[X, Y k0 +1 ] = [X, Y k0 ]Y + Y k0 [X, Y ] =

= k0 (H + k0 ’ 1)Y k0 + Y k0 H = [(k0 + 1)H + k0 (k0 + 1)]Y k0

as desired. Here we used that [H, Y k0 ] = ’2k0 Y k0 .

Now we can proceed with proof of the Lemma. Since H m u = 0, we

see that u ∈ V (0) . Also we notice that Y k u = 0 for some k. Assume

that k ≥ 2. The equality [X, Y k ]u = 0 and the above formula imply

that (H + k ’ 1)(Y k’1 u) = 0. It is easy to see that Y k’1 u ∈ V 2’2k ;

hoewever we also get Y k’1 u ∈ V 1’k . Because k = 1 we have Y k’1 u = 0

and we continue this process to see that Y u = 0 and therefore Hu =

[X, Y ]u = 0 as well.

Let V (») be “the generalized eigenspace” for the eigenvalue » of the

operator H. By de¬nition v ∈ V (») if and only if (H ’ »)k v = 0 for

some natural number k. A vector v ∈ V (») is said to be a vector of

generalized weight ». If furthermore we have Hv = »v we say that

v is of weight ». The space V (») could a priori be bigger than the

eigenspace V » , which is the kernel of H ’ »Id (although we will see

that in fact they are equal for ¬nite-dimensional representations). We

have the decomposition

V = •»∈C V (») .

The » such that V (») = 0 are called the weights of V . Note that X

maps V (») to V (»+2) and V » to V »+2 . Similarly Y maps V (») to V (»’2)

and V » to V »’2 . We say that a weight » is bigger than a weight µ if

» ’ µ ∈ {1, 2, 3 · · ·}. We will soon prove that any weight is always an

integer. If V is a ¬nite-dimensional representation it contains a vector

v of maximal weight ». Then we must have X · v = 0. Let v be a vector

of maximal weight » in V . We claim that » must be an integer. Indeed

let k be the smallest integer such that Y k · v = 0. Then v is killed by

122 CHAPTER 3. HODGE THEORY

[X, Y k ] = k(H + k ’ 1)Y k’1 . Since Y k’1 v is of weight » ’ 2k + 2, it

follows that (H + k ’ 1)Y k’1 · v = (» ’ k + 1)Y k’1 v. Therefore we have

» = k ’ 1. We claim that if V is irreducible then H is diagonalizable,

i.e., V (») = V » for all ». Indeed the eigenvectors of H span an sl(2, C)-

submodule which is non-trivial, hence must be equal to V . It is then

easy to construct a linear map φ : Vk’1 ’ V which sends v0 to v, and

which commutes with the operators H and Y . We then observe

LEMMA 3.6.15 φ is an sl(2, C)-linear map.

Proof. This follows immediately from Lemma 3.6.14.

This proves Theorem 3.6.12.

Now given any ¬nite-dimensional representation V we can ¬nd a

Jordan-H¨lder series

o

0 ‚ E1 ‚ · · · ‚ Ek ‚ V,

where E1 , ..., Ek are sl(2, C)-submodules, such that this ¬ltration is pre-

served by H, X, and Y ; and E1 , E2 /E1 , ..., V /Ek all are irreducible

representations. The di¬culty is to show that E2 ’E1 •(E2 /E1 ), etc...

˜

LEMMA 3.6.16 Let

0’V ’W ’C’0

be an extension of sl(2, C) ¬nite-dimensional representation V . Then

W = V • C as an sl(2, C)-representation.

Proof. We observe that X : V (0) ’ V (2) is surjective, for irreducible

representations. It follows that the same is true for any ¬nite-dimensional

representation, because given an exact sequence 0 ’ E ’ F ’ G ’ 0

of sl(2, C)-representations, we have a diagram with exact rows:

0 ’ E (0) ’ F (0) ’ G(0) ’ 0

“X “X “X .

0 ’ E (2) ’ F (2) ’ G(2) ’ 0

¨

3.6. HODGE THEORY ON KAHLER MANIFOLDS 123

We have

onto

Ker(X on W ) ’ C

“ “

0 ’ V (0) W (0)

’ ’ ’0

C

“X “X “X

0 ’ V (2) W (2)

’ ’ 0 ’0

“ “

0 0

The fact that the kernel of the mapping X : W (0) ’ W (2) maps onto C

means that there exists u ∈ W such that Xu = 0, u ∈ W (0) , and u maps

to 1 ∈ C. By the previous Lemma we have that u generates a trivial

1-dimensional sl(2, C) representation C · u, and therefore W = V • C · u.

We are now ready to prove

PROPOSITION 3.6.17 Let 0 ’ V1 ’ V2 ’ V3 ’ 0 be an ex-

act sequence of ¬nite-dimensional representations of sl(2, C). Then the

extension is split, i.e., V3 ’V1 • V2 as sl(2, C)-representations.

˜

Proof. Given representations A and B of sl(2, C), the space Hom(A, B)

becomes a representation by the rule (Xφ)(a) = X · φ(a) ’ φ(X · a) for

φ ∈ Hom(A, B), a ∈ A. Then the space E of linear maps f : V2 ’ V1

such that f (v) = »v for all v ∈ V1 is a subrepresentation of Hom(V2 , V1 )

and we have an exact sequence

»

0 ’ Hom(V3 , V1 ) ’ E ’ C ’ 0.

The previous lemma implies that the extension splits. This means that

there exists an element f of E which is killed by sl(2, C) and induces

the identity on V1 ‚ V2 . The fact that u is killed by sl(2, C) precisely

means that u is a homomorphism of sl(2, C)-representations. Then u is

onto, and if K is its kernel we have a decomposition V2 = V1 • K of

sl(2, C)-representations, where K ’V3 . This proves the splitting.

˜

Now starting with a Jordan-H¨lder series 0 ‚ E1 ‚ · · · ‚ Ek ‚ V

o

for a representation V we see inductively that Ej is a direct sum of

irreducible representations. This proves Theorem 3.6.13. As another

consequence one obtains

124 CHAPTER 3. HODGE THEORY

COROLLARY 3.6.18 For any ¬nite-dimensional representation V

of

sl(2, C) the operator H is diagonalizable and the map

X j : V ’j ’ V j , j > 0

is an isomorphism.

Proof. We can assume that V is an irreducible representation of sl(2, C):

V Vm . Then using the explicit structure of Vm = Pm we see that

when j > m then V ’j = V j = 0 and when j ¤ m the action of

m’j m+j

‚

X j = (x ‚y )j is non zero on the monomial x 2 y 2 in Pm .

Since H is diagonalizable, we also point out that V (») = V » .

THEOREM 3.6.19 (Hard Lefschetz theorem.) Let M be a com-

pact K¨hler manifold of dimension n. Then the map

a

Lk : H n’k (M, C) ’ H n+k (M, C), k ≥ 0

is an isomorphism.

Proof. We use the isomorphism H — (M, C) H— (M ) together with the

correspondence X ” L, Y ” Λ, and H ” P to view H— (M ) as a

representation space of sl(2, C). On (n ’ k)-forms P acts as ’k and

thus the result follows from the following commutative diagram.

Lk

n’k

(M ) ’’ Hn+k (M )

H

|| ||

Xk

V ’k Vk

’’

COROLLARY 3.6.20 If p is odd then dim H p (M, R) is an even num-

ber.

¨

3.6. HODGE THEORY ON KAHLER MANIFOLDS 125

Proof. We will give two proofs. The ¬rst is simple and the second is